## Calculus with Applications (10th Edition)

$$r'\left( t \right) = \frac{{ - 15{t^2} + 52t - 7}}{{{{\left( {3t + 1} \right)}^4}}}$$
\eqalign{ & r\left( t \right) = \frac{{5{t^2} - 7t}}{{{{\left( {3t + 1} \right)}^3}}} \cr & {\text{differentiate both sides}} \cr & r'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{5{t^2} - 7t}}{{{{\left( {3t + 1} \right)}^3}}}} \right] \cr & {\text{use the quotient rule}} \cr & r'\left( t \right) = \frac{{{{\left( {3t + 1} \right)}^3}\frac{d}{{dt}}\left[ {5{t^2} - 7t} \right] - \left( {5{t^2} - 7t} \right)\frac{d}{{dt}}\left[ {{{\left( {3t + 1} \right)}^3}} \right]}}{{{{\left( {{{\left( {3t + 1} \right)}^3}} \right)}^2}}} \cr & {\text{use the chain rule }}\frac{d}{{dx}}\left[ {g{{\left( x \right)}^n}} \right] = ng{\left( x \right)^{n - 1}}\frac{d}{{dx}}\left[ {g'\left( x \right)} \right]{\text{ then}} \cr & r'\left( t \right) = \frac{{{{\left( {3t + 1} \right)}^3}\frac{d}{{dt}}\left[ {5{t^2} - 7t} \right] - \left( {5{t^2} - 7t} \right)\left( 3 \right){{\left( {3t + 1} \right)}^2}\frac{d}{{dt}}\left[ {3t + 1} \right]}}{{{{\left( {3t + 1} \right)}^6}}} \cr & {\text{find derivatives}} \cr & r'\left( t \right) = \frac{{{{\left( {3t + 1} \right)}^3}\left( {10t - 7} \right) - \left( {5{t^2} - 7t} \right)\left( 3 \right){{\left( {3t + 1} \right)}^2}\left( 3 \right)}}{{{{\left( {3t + 1} \right)}^6}}} \cr & {\text{simplifying}} \cr & r'\left( t \right) = \frac{{\left( {3t + 1} \right)\left( {10t - 7} \right) - 9\left( {5{t^2} - 7t} \right)}}{{{{\left( {3t + 1} \right)}^4}}} \cr & r'\left( t \right) = \frac{{30{t^2} - 21t + 10t - 7 - 45{t^2} + 63t}}{{{{\left( {3t + 1} \right)}^4}}} \cr & r'\left( t \right) = \frac{{ - 15{t^2} + 52t - 7}}{{{{\left( {3t + 1} \right)}^4}}} \cr}